Prove that $f\in \mathcal{L}^1$ iff $\int|f|d\mu<\infty$

Question

One of the properties of Lebesgue integrable functions, as stated here problem 3.7, is $f\in \mathcal{L}^1$ iff $\int|f|d\mu<\infty$ ,where $\mathcal{L}^1$ is the family of all integrable functions.

I tried proving this by: $$f\in \mathcal{L}^1\implies \int f^+d\mu<\infty \text{ and } \int f^-d\mu<\infty$$ hence $$\int |f|d\mu=\int |f^+-f^-|d\mu$$

And here I don't know how can I break up the integral. on the other hand: $$\int |f|d\mu<\infty\implies\int |f|d\mu=\int |f|^+d\mu-\int |f|^-d\mu=\int |f|^+d\mu=\int f^+d\mu<\infty$$ Is this any good?

How do I show both sides?

Generally the definition of $\mathcal{L}^1$ is $\int |f|<\infty$. — carmichael561, Oct 02 '16 at 23:48
You could make your question better by providing all the definitions of the key terms in your statement. — , Oct 03 '16 at 00:26
Then one might ask what is your definition of "Lebesgue integrable". Different textbooks would probably give different definitions. Without giving appropriate context, it would be difficult for others to answer your questions. — , Oct 03 '16 at 00:33
@Jack I have edited to include the source of my question as an overkill. — havakok, Oct 03 '16 at 00:38

score 1 · Accepted Answer · answered Oct 03 '16 at 10:49

By definition, $$f \in \mathcal L^1 \quad \text{ iff } \quad \int f^+ d\mu < \infty \text{ and } \int f^- d\mu < \infty.$$ Now assume $f \in \mathcal L^1$. Since $|f| = f^+ + f^-$, we get $$\int |f| d\mu = \int f^+ + f^- d\mu = \int f^+ d\mu + \int f^- d\mu$$ by linearity, and both of these are by definition $< \infty$, so $\int |f|d\mu < \infty$.

For the opposite direction, we will prove the contrapositive. Assume that $f \notin \mathcal L^1$. Again we get $$\int|f|d\mu = \int f^+ d\mu + \int f^- d\mu,$$ but since $f \notin \mathcal L^1$, one of two summands on the right side must be $\infty$. Hence $\int |f|d\mu = \infty$.

Prove that $f\in \mathcal{L}^1$ iff $\int|f|d\mu<\infty$

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